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Sklearn: Pass class names to make_scorer


Sklearn: Pass class names to make_scorer

By : tashfeen
Date : October 24 2020, 08:10 PM
I think the issue was by ths following , You can use custom scoring method described here in user guide where the signature is:
code :
func(estimator, X, y)
def accuracy_2_cv(estimator, X, y_labels):
    n=2
    y_probs = estimator.predict_proba(X)
    class_names = estimator.classes_
    cons_preds = [top_n_consolidation(y_labels[i], y_probs[i,:], class_names, n) for i in range(y_probs.shape[0])]
    return accuracy_score(y_true=y_labels, y_pred=cons_preds)
custom_scoring = {'acc'       : 'accuracy',
                  'acc2'      : accuracy_2_cv}


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(Python - sklearn) How to pass parameters to the customize ModelTransformer class by gridsearchcv

(Python - sklearn) How to pass parameters to the customize ModelTransformer class by gridsearchcv


By : user2984518
Date : March 29 2020, 07:55 AM
will help you GridSearchCV has a special naming convention for nested objects. In your case ess__rfc__n_estimators stands for ess.rfc.n_estimators, and, according to the definition of the pipeline, it points to the property n_estimators of
code :
ModelTransformer(RandomForestClassifier(n_jobs=-1, random_state=1,  n_estimators=100)))
parameters = {
  'ess__rfc__model__n_estimators': (100, 200),
}
make_scorer giving an error: 'str' object is not callable make_scorer

make_scorer giving an error: 'str' object is not callable make_scorer


By : user997016
Date : March 29 2020, 07:55 AM
it helps some times I am using the following code to optimize a random forest algorithm but this is throwing a typeerror: 'str' object is not callable. Could you please help me in identifying what could be the cause? The fit line is throwing the error in the scorer.py file for this one "score = scorer(estimator, X_test, y_test)" , Change your third line to:
code :
clf_scorer = make_scorer(roc_auc_score)
How to use make_scorer Custom scoring function in sklearn

How to use make_scorer Custom scoring function in sklearn


By : Lesiba Bothitho Mara
Date : March 29 2020, 07:55 AM
To fix the issue you can do The solution is in adding a parameter called needs_proba=True in the make_scorer function! This works ok.
code :
def top_decile_conversion_rate(y_prob, y_actual):
    # Function goes in here
    print "---prob--"
    print y_prob
    print "---actual--"
    print y_actual
    print "---end--"

    return 0.5


features = pd.DataFrame({"f1":np.random.randint(1,1000,500) , "f2":np.random.randint(1,1000,500), 
                         "label":[round(x) for x in np.random.random_sample(500)]})


my_scorer = make_scorer(top_decile_conversion_rate, greater_is_better=True,needs_proba=True)
gs = grid_search.GridSearchCV(
    estimator=LogisticRegression(),
    param_grid={'C': [i for i in range(1, 3)], 'class_weight': [None], 'penalty':['l2']},
    cv=20,
    scoring=my_scorer ) 
model = gs.fit(features[["f1","f2"]], features.label)
Type Error with sklearn make_scorer Function

Type Error with sklearn make_scorer Function


By : Carlo Luis
Date : March 29 2020, 07:55 AM
With these it helps I believe 3 positional arguments are being passed since you called the method on an instance. Try adding self as the first param to that method.
code :
def make_custom_score(self, y_true, y_score):
Pass estimator to custom score function via sklearn.metrics.make_scorer

Pass estimator to custom score function via sklearn.metrics.make_scorer


By : 108
Date : March 29 2020, 07:55 AM
seems to work fine There is an alternative way to make a scorer mentioned in the documentation. Using this method I can do the following:
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